End of Chapter Exercises
Exercise 1:
Problem 1:
The stomach secretes gastric juice, which contains hydrochloric acid. The gastric juice
helps with digestion. Sometimes there is an overproduction of acid, leading to heartburn or
indigestion. Antacids, such as milk of magnesia, can be taken to neutralise the excess acid.
Milk of magnesia is only slightly soluble in water and has the chemical formula Mg(OH)2.
1. Write a balanced chemical equation to show how the antacid reacts with the acid.
2. The instructions on the bottle recommend that children under the age of 12 years take
one teaspoon of milk of magnesia, whereas adults can take two teaspoons of the
antacid. Briefly explain why the dosages are different.
3. Why is it not advisable to take an overdose of the antacid? Refer to the hydrochloric
acid concentration in the stomach in your answer.
(DoE Grade 11 Exemplar, 2007)
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Answer 1:
1. base + acid → salt + water
The cation (from the base) is Mg2+. The anion (from the acid) is Cl−.
Therefore the salt is MgCl2
Mg(OH)2(s)+2HCl(aq) → MgCl2(aq)+2H2O(ℓ)
2. Adults have a bigger mass and generally produce more acid than children. Adults will
therefore need more antacid to neutralise the excess acid. If children were to take the
same dosage as adults they would have excess base in their stomach.
3. A low acid concentration (pH too high) in the stomach may slow down food digestion or
may cause further stomach upset.
Problem 2:
The compound NaHCO3 is commonly known as baking soda. A recipe requires 1,6 g of
baking soda, mixed with other ingredients, to bake a cake.
1. Calculate the number of moles of NaHCO3 used to bake the cake.
2. How many atoms of oxygen are there in 1,6 g of baking soda?
3. During the baking process, baking soda reacts with an acid (e.g. acetic acid in vinegar)
to produce carbon dioxide and water, as shown by the reaction equation below:
NaHCO3(aq)+CH3COOH(aq) → CH3COONa(aq)+CO2(g)+H2O(ℓ)
Use the above equation to explain why the cake rises during this baking process.
(DoE Grade 11 Paper 2, 2007)
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Answer 2:
1. M(NaHCO3) = 23,0 + 1,01 + 12,0 + (3 x 16,0) = 84,01 g.mol−1
n(NaHCO3) = mM=1,6 g84,01 g.mol−1=0,019 mol
2. NA (Avogadro's number) = 6,022 × 1023 atoms.mol−1.
Number of atoms = n x NA = 0,019 mol x 6,022 × 1023 atoms.mol−1
Number of atoms = 1,14 × 1022 atoms.
3. The carbon dioxide gas that is formed during the reaction forms bubbles in the cake
mixture causing it to rise along with the expanding volume of the gas bubbles.
Problem 3:
Label the acid-base conjugate pairs in the following equation:
HCO−3(aq)+H2O(ℓ) ⇌ CO2−3(aq)+H3O+(aq)
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Answer 3:
Problem 4:
A certain antacid tablet contains 22,0 g of baking soda (NaHCO3). It is used to neutralise
the excess hydrochloric acid in the stomach. The balanced equation for the reaction is:
NaHCO3(s)+HCl(aq) → NaCl(aq)+H2O(ℓ)+CO2(g)
The hydrochloric acid in the stomach has a concentration of 1,0 mol.dm −3. Calculate the
volume of the hydrochloric acid that can be neutralised by the antacid tablet.
(DoE Grade 11 Paper 2, 2007)
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Answer 4:
M(NaHCO3) = 23,0 + 1,01 + 12,0 + (3 x 16,0) = 84,01 g.mol−1
n(NaHCO3)=mM
n=22,0 g84,01 g.mol−1 = 0,26 mol
From the balanced equation we see that the molar ratio of NaHCO3 to HCl is 1:1.
Therefore, n(HCl) = n(NaHCO3) = 0,26 mol
C=nV therefore V=nC
V(HCl)=0,26 mol1,0 mol.dm−3 = 0,26 dm3
Problem 5:
A learner finds some sulfuric acid solution in a bottle labelled 'dilute sulfuric acid'. He wants
to determine the concentration of the sulfuric acid solution. To do this, he decides to titrate
the sulfuric acid against a standard potassium hydroxide (KOH) solution.
1. What is a standard solution?
2. Calculate the mass of KOH which he must use to make 300 cm 3 of a 0,2
mol.dm−3 KOH solution.
3. Calculate the pH of the 0,2 mol.dm−3 KOH solution (assume standard temperature).
4. Write a balanced chemical equation for the reaction between H2SO4 and KOH.
5. During the titration the learner finds that 15 cm 3 of the KOH solution neutralises 20
cm3 of the H2SO4solution. Calculate the concentration of the H2SO4 solution.
(IEB Paper 2, 2003)
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Answer 5:
1. A standard solution is a solution that contains a precisely known concentration of a
substance. This substance can then be used in titrations.
2. V(KOH) = 300 cm3 ×0,001 dm31 cm3 = 0,3 dm3
C=nV, therefore n = C x V
n(KOH) = 0,2 mol.dm−3 x 0,3 dm3 = 0,06 mol
M(KOH) = (39,1 + 16,0 + 1,01) g.mol−1 = 56,11 g.mol−1
n=mM, therefore m = n x M
m(KOH) = 0,06 mol x 56,11 g.mol−1 = 3,37 g
3. KOH is a very strong base and will completely dissociate in water. Therefore the
concentration of OH−ions is the same as the concentration of the solution.
pH is defined as: pH = -log[H+]
However, we can also define pOH in a similar way: pOH = -log[OH−]
The relationship between the two is: 14 = pH + pOH, therefore pH = 14 - pOH
pH = 14 - (-log[OH−]) = 14 - (-log[0,2]) = 14 - 0,7 = 13,3
4. Sulfuric acid is a strong acid and potassium hydroxide is a strong base, therefore the
equation is:
H2SO4(aq)+2KOH(aq) → K2SO4(aq)+2H2O(ℓ)
5. V(KOH) = 15 cm3 ×0,001 dm31 cm3 = 0,015 dm3
The number of moles of KOH used to neutralise the H2SO4 is:
n(KOH) = C x V = 0,2 mol.dm−3 x 0,015 dm3 = 0,003 mol
From the balanced equation we see that the mole ratio of H2SO4 to KOH is 1:2. There
is one mole of sulfuric acid for every two moles of potassium hydroxide.
n(H2SO4) = 0,003 mol2 = 0,0015 mol
V(H2SO4) = 20 cm3 ×0,001 dm31 cm3 = 0,020 dm3
Therefore, the concentration of the sulfuric acid is:
C(H2SO4)=nV=0,0015 mol0,020 dm3 = 0,075 mol.dm−3